# Given equilateral triangle with vertex z1=1, z2=2+i, find z3 complex?

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### 2 Answers

The problem provides the information that the triangle whose vertices are the complex numbers `z_1,z_2,z_3` is equilateral, hence, all its three sides are equal.

Evaluating the lengths of the sides yields:

`|z_2 - z_1| = |z_3 - z_1| = |z_3 - z_2|`

`|2 + i - 1| = |x + i*y - 1| = |x + i*y - 2 - i|`

`|1 + i| = |x - 1 + i*y| = |x - 2 + i*(y - 1)|`

Evaluating the modules of complex nmbers yields:

`sqrt(1^2 + 1^2) = sqrt((x - 1)^2 + y^2) = sqrt((x - 2)^2 + (y - 1)^2)`

`{(sqrt 2 = sqrt((x - 1)^2 + y^2)),(sqrt 2 = sqrt((x - 2)^2 + (y - 1)^2)):}`

`{(2 = (x - 1)^2 + y^2),(2 = (x - 2)^2 + (y - 1)^2):}`

`{(2 = x^2 - 2x + 1 + y^2),(2 = x^2 - 4x + 4 + y^2 - 2y + 1):}`

`{(x^2 - 2x + y^2 = 1),(-3 = x^2 - 4x + y^2 - 2y):}`

Replacing `1 + 2x` for `x^2+ y^2` yields:

`{(x^2 - 2x + y^2 = 1),(-3 = 1 + 2x - 4x - 2y):}`

`{(x^2 - 2x + y^2 = 1),(-4 = -2x - 2y):} => {(x + y = 2),(x^2 - 2x + y^2 = 1):}`

`y = 2 - x => x^2 - 2x + (2 - x)^2 = 1 => x^2 - 2x + 4 - 4x + x^2 = 1`

`2x^2 - 6x + 3 = 0`

Using quadratic formula yields:

`x_(1,2) = (6+-sqrt(36 - 24))/4 => x_(1,2) = (6+-sqrt(12))/4`

`x_(1,2) = (6+-2sqrt(3))/4 => x_(1,2) = (3+-sqrt(3))/2`

`y = 2 - (3+-sqrt(3))/2 => y = (1+-sqrt(3))/2`

**Hence, evaluating the complex number z_3, under the given conditions, yields: **`z_3 = (3+-sqrt(3))/2 + (1+-sqrt(3))/2*i.`

Actually the point are two, now let we conisder every complex number as a point in the Real plane. So the complex `z_1=1` is actually the point `P(1;0)` , and the complex `z_2=2+i` the point `Q(2;1)` .

So the line between `P ` and `Q` is a side of equilater triangle.

`s=sqrt((2-1)^2+(1-0)^2)=sqrt(2)`

The middle point is `M(3/2;1/2)`

Calculating equation of the straight line `r` passing by `P` and `Q` :

`y-y_0=m(x-x_0)`

So: `1-0=m(2-1) rArr m=1`

Subduing passage for `P` or `Q` , doesn't matter,we obtain staright line equation:

`y=x-1`

Now a perpendicular line , passing by middle point between `P` and `Q` stands for height `h` of equilater triangle, that we know:

`h=sqrt(3)/2 s`

since `s=sqrt(2) rArr h=sqrt(3)/sqrt(2)`

Now the symmetric points `R` and `S` ,lyng on a straight line `t` ,perepndicular at `r` and colliding it a the middle point `M` so that the distance between `R` and `M ` and between `S` and `M ` is equal at `sqrt(3)/sqrt(2)` , that is height of triangle value,are the last vertex we are searching for.

Calucating `t` equation:

Being perpendicular at `r` , the value of its slope `m'` is: `mm'=-1` (`m ` slope of `r` ).

Then `m'=-1`

`y=-x+q`

since `M` lies on `t` we find: `y= -x+2` as `t ` equation.

The points `R` and `S` accomplish then this two conditions:

1) Both lay on line `t`

2) Their distance from `M` (and so from the line `r` ) is to be `sqrt(3)/sqrt(2)`

Using distance point `(x_0;y_0)` from straight line: `ax+by+c=0` formula:

`d= (|ax_0+by_0+c|)/sqrt(a^2+b^2)`

Re -write `r ` equation as: `x-y-1=0` we get:

`(x_0-y_0-1)/(+-sqrt((1)^2+(-1)^2))= sqrt(3)/sqrt(2)` `rArr` `(x_0-y_0-1)/(+-sqrt(2))=sqrt(3)/sqrt(2)`

Realy we have got two equations:

`x_0-y_0-1=sqrt(3)`

`x_0-y_0-1=-sqrt(3)`

Since both lay on `t` `y_0=-x_0+2`

Substituing:

`2x_0-3=sqrt(3)`

`2x_0-3=-sqrt(3)`

So: `x_0=(3+sqrt(3))/2` `y_0=(1-sqrt(3))/2`

and `x_0=(3-sqrt(3))/2` `y_0=` `(1+sqrt(3))/2`

So the third vertex are both the complex numbers:

`z_3=1/2 [3+sqrt(3)+i(1-sqrt(3))]`

`z_4=1/2[3-sqrt(3)+i(1+sqrt(3))]`